一道高数题设f’’(x)连续,且f(∏)=0,∫(上限∏,下限0)f(x)+f’’(x)sinxdx=5,求f(0)

Question

一道高数题
设f’’(x)连续,且f(∏)=0,∫(上限∏,下限0)【f(x)+f’’(x)】sinxdx=5,求f(0)

Answer 1

将两个分开,分别用分部积分：
5=∫(0－∏)[f(x)+f''(x)]sinxdx
=∫(0－∏)f(x)sinxdx+∫(0－∏)f''(x)sinxdx
=-∫(0－∏)f(x)dcosx+∫(0－∏)sinxdf'(x)
=-f(x)cosx|(0－∏)+∫(0－∏)cosxdf(x)+sinxf'(x)|(0－∏)-∫(0－∏)f'(x)dsinx (第一项和第四项消去）
=-f(x)cosx|(0－∏)+sinxf'(x)|(0－∏)
=-f(∏)cos∏+f(0)cos0
=f(0)
f(0)=5